SOLUTION: The product of two consecutive integers is 71 more than their sum. Find the integers. I think that the integers will be 71 plus the total of the two integers but I am really con

Question 242807: The product of two consecutive integers is 71 more than their sum. Find the integers.
I think that the integers will be 71 plus the total of the two integers but I am really confused on how you will get that.
Answer by oberobic(2304) (Show Source): You can put this solution on YOUR website!
Setup what you know using the fewest unknowns.
Two consecutive numbers could be called x & y.
BUT, calling them x and x+1 is much easier and uses only one unknown.
The product would be:
x(x+1) = x^2+ x
The sum would be:
x + (x+1) = 2x + 1
...
We are told the product is 71 more than the sum, or the sum + 71 = product.
Substituting the formulas we have developed
2x + 1 + 71 = x^2 + x
...
Now we need to do the algebra to get it into standard form...
Subtracting x^2 + x from both sides
2x + 1 + 71 - (x^2 + x) = 0
Simplify
2x + 1 + 71 - x^2 - x = 0
Rearranging and combining terms
-x^2 + x + 72 = 0
Multiplying both sides by -1 to make it look friendlier...
x^2 - x - 72 = 0
Factoring we arrive at
(x + 8)(x - 9) = 0
So x must be either -8 or 9.
(Actually, algebra tells us that BOTH are answers.)
...
Checking our work is now the key to knowing the answer.
If x = 9, then the two consecutive integers are 9 and 10.
Does 2x + 1 + 71 = x^2 + x, when x = 9?
2(9) + 1 + 71 (equals?) 9(9) + 9??
18 + 1 + 71 = 18 + 72 = 9(2) + 9(8) = 9(10) = 90
9(9) + 9 = 81 + 9 = 90
So x = 9 works quite nicely.
...
What about x = -8?
(ugly grunt work must be done.)
Does 2x + 1 + 71 = x^2 + x, when x = -8?
2(-8) + 1 + 71 = -16 + 1 + 71 = -16 + 72 = 56
(-8)^2 + (-8) = 64 - 8 = 56
56 = 56 is TRUE! (smile)
...
We have two answers.
x = -8 or 9
The two consecutive numbers are either 8 & 9 or -8 & -7.
...
Note that we are not too surprised by having two solutions because we have a QUADRATIC equation.
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